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\(D=50^2-49.51\)
\(\Leftrightarrow D=50^2-\left(50-1\right)\left(50+1\right)\)
\(\Leftrightarrow D=50^2-50^2+1=1\)
\(C=39^2+78.61+61^2\)
\(\Leftrightarrow C=39^2+2.39.61+61^2\)
\(\Leftrightarrow C=\left(39+61\right)^2=100^2=10000\)
D=(502-492)+(482-472)+...+(22-12)
= ( (50-49)(50+49)+(48-47)(48+47)+...+(2-1)(2+1)
= 50+49+48+47+...+2+1
=(50+1).502
=1275
cho mik sửa tí\(\frac{\left(50+1\right)x2}{50}\)nhé
___________________________
_chúc bạn học tốt_
C=502 - 492 +482 -472 +...+22 -12
=(502 - 492)+(482 -472 )+.....+(22-1)(
= (50 - 49)(50 + 49) + (48 – 47)(48 + 47) + ... +
(2 + 1)(2 – 1)
=(50+49).1+(48+47).1+.....+(2+1).1
= 50 + 49 + 48 + 47 + ... + 2 + 1
= (50 + 1) + (49 + 2) + ... + (25 +26)
= 51 . 25 = 1275
Áp dụng HĐT a2 - b2 = ( a + b )( a - b ) ta có :
502 - 492 + 482 - 472 + ... + 22 - 12
= ( 502 - 492 ) + ( 482 - 472 ) + ... + ( 22 - 12 )
= ( 50 + 49 )( 50 - 49 ) + ( 48 + 47 )( 48 - 47 ) + ... + ( 2 + 1 )( 2 - 1 )
= 99.1 + 95.1 + ... 3.1
= 99 + 95 + ... + 3
= \(\frac{\left(99+3\right)\left[\left(99-3\right):4+1\right]}{2}\)
= 1275
\(M=31^2+2.31.19+19^2\)
\(\Rightarrow M=\left(31+19\right)^2\)
\(\Rightarrow M=50^2\)
\(\Rightarrow M=2500\)
\(N=45^2-90.35+25^2\)
\(\Rightarrow N=45^2-2.45.35+25^2\)
\(\Rightarrow N=\left(45-25\right)^2\)
\(\Rightarrow N=20^2=400\)
\(P=51^2-50^2+49^2-48^2+...+3^2-2^2+1^2\)
\(\Rightarrow P=\left(51-50\right)\left(51+50\right)+\left(49-48\right)\left(49+48\right)+...+\left(3-2\right)\left(3+2\right)+1\)
\(\Rightarrow P=101+97+...+5+1\)
\(\Rightarrow P=\frac{\left(101+1\right)\left[\left(101-1\right):2+1\right]}{2}\)
\(\Rightarrow P=102.51:2=51.51=51^2\)
Bài 2:
Ta có: \(A=\left(x^2-3x+1\right)\left(x^2-3x-1\right)=x^4-1\) > -1
=> Bmin = -1 <=> \(x^4-1=-1=>x=0\)
vậy Bmin= 1 <=> \(x=0\)
\(\left(8x^3-60x^2+150x-125\right)-\left(27x^3-108x^2+144x-64\right)+\left(x^3+3x^2+3x+1\right)=0\)
\(-18x^3+51x^2+9x-60=0\)
\(\left(2x-5\right)\left(x+1\right)\left(3x-4\right)=0\)
\(\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-1\\x=\frac{4}{3}\end{array}\right.\)